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प्रश्न
An observer, 1.5m tall, is 28.5m away from a tower 30m high. Determine the angle of elevation of the top of the tower from his eye.
बेरीज
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उत्तर
Here, ED is the height of the observer and AC is the tower.
BE = CD = 28. 5 m
AB = AC - BC = 30 m - 1.5 m = 28.5 m
In ΔABE,
`tan <"ABE" = "AB"/"BE"`
⇒ `tan <"ABE" = (28.5"m")/(28.5"m") = 1`
But, `tan 45^circ = 1`
`therefore <"ABE" = 45^circ`
Thus , the required angle of elevation is `45^circ`,
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